energyenergy-conservationnewtonian-mechanicsrotationrotational-dynamics
Why is it not necessary to take into account rotational kinetic energy when using the Law of Conservation of Mechanical Energy to solve vertical circular motion problems? After all, the particle is rotating about the centre of the circle and does have rotational KE, doesn't it?
All the examples I have seen just use KE= 1/2 $mv^2$, e.g. here: http://www.physicsforums.com/showpost.php?p=2312566&postcount=4
Your answer seems to be good. I solved it in the same way. However there are two points in which i have no clarity.
The first thing i do not have clear is the conservation of energy in the problem. We use the energy conservation, but how do we guarantee that the normal force is conservative? And, still, in the conservation equations we do not consider its potential energy. I would thank some guide in this point.
The other one is if it is necessary to consider the translational energy. I think that it is not. I thought about it in terms of the physical pendulum. In this case the center of mass is moving, but to get the correct equation of motion from the energy conservation, I do not consider the translational energy of the CM.
I came up with a solution seeing John Rennie's comment.
The centripetal force, $\vec F= -F \hat r $ so infinitesimal work done by centripetal force, $$dW=\vec F.d \vec r= -F \hat r.d\vec r$$
but, $\hat r⊥d \vec r$ so $$dW=0$$ 
is this correct ?
Best Answer
For a point particle, Translational KE is rotational KE:
$$\frac12I\omega^2=\frac12mr^2\omega=\frac12mv^2$$
The formula for rotational KE ($\frac12I\omega^2$) is derived by adding up the KE of each particle in a rigid body in pure rotation. When body has both rotation and translation, we can derive that:
$$KE=KE_R+KE_T=\frac12I_{com}\omega_{com}^2+\frac12mv_{com}^2$$
In this case, the point particle has no $\omega$ about its center of mass, so no problem. Though we can still apply the pure rotation formula.
Just that we can't say it has both rotational and translational mortion.